ACM: 条件最短路 poj 1724 (没剪枝…

                                                                  
ROADS

Description

N cities named with numbers 1 ... N are connected with one-way
roads. Each road has two parameters associated with it : the road
length and the toll that needs to be paid for the road (expressed
in the number of coins).

Bob and Alice used to live in the city 1. After noticing that Alice
was cheating in the card game they liked to play, Bob broke up with
her and decided to move away - to the city N. He wants to get there
as quickly as possible, but he is short on cash.

We want to help Bob to find the shortest path from
the city 1 to the city N that he can afford with
the amount of money he has.

Input

The first line of the input contains the integer K, 0
<= K <= 10000, maximum number of
coins that Bob can spend on his way.

The second line contains the integer N, 2 <= N
<= 100, the total number of cities.

The third line contains the integer R, 1 <= R
<= 10000, the total number of roads.

Each of the following R lines describes one road by specifying
integers S, D, L and T separated by single blank characters
:

S is the source city, 1 <= S <=
N

D is the destination city, 1 <= D
<= N

L is the road length, 1 <= L <=
100

T is the toll (expressed in the number of coins), 0
<= T <=100

Notice that different roads may have the same source and
destination cities.

Output

The first and the only line of the output should contain the
total length of the shortest path from the city 1 to the city N
whose total toll is less than or equal K coins.

If such path does not exist, only number -1 should be written to
the output.

Sample Input

5

6

7

1 2 2 3

2 4 3 3

3 4 2 4

1 3 4 1

4 6 2 1

3 5 2 0

5 4 3 2

Sample Output

11

 

题意: 现在要从1 到达 n城市, 每个城市之间的链接都知道, 道路的长度和费用.

        
求在小于等于费用K下最短路径.

解题思路:

               
1. 条件最短路. 依然是记忆化深搜 + 剪枝.

               
2. 可恨的测试数据有N多个环. TLE多次.

代码:

#include <cstdio>

#include <iostream>

#include <cstring>

using namespace std;

#define MAX 1005

const int INF = 1 << 30;

struct node

{

    int v;

    int
len;

    int
cost;

    int
next;

}edges[MAX*10];

int K;

int n, m;

int next[MAX*10];

int vis[MAX];

int num;

int result;

void init()

{

    num =
0;

    result =
INF;

  
 memset(next,-1,sizeof(next));

  
 memset(vis,0,sizeof(vis));

}

void dfs(int v,int w,int cost)

{

    if(w
> INF)

  
   
 return ;

    if(v ==
n)

    {

  
   
 if(result > w)

  
   
   
 result = w;

  
   
 return ;

    }

    

    for(int e =
next[v]; e != -1; e = edges[e].next)

    {

  
   
 int len = edges[e].len;

  
   
 int t_cost = edges[e].cost;

  
   
 if( vis[ edges[e].v ] <= 1)

  
   
 {

  
   
   
 vis[ edges[e].v ]++;

  
   
   
 if(t_cost + cost <= K
&& w+len <
result)

  
   
   
   
 dfs(edges[e].v,w+len,t_cost+cost);

  
   
   
 vis[ edges[e].v ]--;

  
   
 }

    }

}

int main()

{

//  
 freopen("input.txt","r",stdin);

    int u, v,
len, cost;

  
 while(scanf("%d %d
%d",&K,&n,&m) !=
EOF)

    {

  
   
 num = 0;

  
   
 result = INF;

  
   
 memset(next,-1,sizeof(next));

  
   
 memset(vis,0,sizeof(vis));

  
   
 for(int i = 1; i <= m; ++i)

  
   
 {

  
   
   
 scanf("%d %d %d
%d",&u,&v,&len,&cost);

  
   
   
 edges[num].v = v;

  
   
   
 edges[num].len = len;

  
   
   
 edges[num].cost = cost;

  
   
   
 edges[num].next = next[u];

  
   
   
 next[u] = num;